Difference between revisions of "2022 AMC 10A Problems/Problem 22"
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<math>\textbf{(A) }4082\qquad\textbf{(B) }4095\qquad\textbf{(C) }4096\qquad\textbf{(D) }8178\qquad\textbf{(E) }8191</math> | <math>\textbf{(A) }4082\qquad\textbf{(B) }4095\qquad\textbf{(C) }4096\qquad\textbf{(D) }8178\qquad\textbf{(E) }8191</math> | ||
+ | == Video Solution 1 By Omega Learn == | ||
+ | https://www.youtube.com/watch?v=gW8gPEEHSfU&list=PLT9bNzqjDoMl3jNviYrczw7Ck_ArS54Xn&index=6 | ||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2022|ab=A|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} | ||
== See Also == | == See Also == |
Revision as of 03:13, 12 November 2022
Contents
Problem
Suppose that 13 cards numbered are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards 1, 2, 3 are picked up on the first pass, 4 and 5 on the second pass, 6 on the third pass, 7, 8, 9, 10 on the fourth pass, and 11, 12, 13 on the fifth pass. For how many of the possible orderings of the cards will the cards be picked up in exactly two passes?
[asy]size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy]
Video Solution 1 By Omega Learn
https://www.youtube.com/watch?v=gW8gPEEHSfU&list=PLT9bNzqjDoMl3jNviYrczw7Ck_ArS54Xn&index=6
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Solution by Omega Learn Using Combinatorial Identities and Overcounting
https://www.youtube.com/watch?v=gW8gPEEHSfU&list=PLT9bNzqjDoMl3jNviYrczw7Ck_ArS54Xn&index=6